How can a photon have no mass if it has momentum?

[jaurandt]

My personal course of study in quantum mechanics hasn't brought me this far and so this question may be incredibly naive, but it has still been troubling me. If the energy of a photon is

E = (hbar)(omega)

and the units of hbar are J*s (obviously), then how can a photon have no mass if a J is defined as 1 kg(m^2)/(s^2)? Is there some kind of limit involved here where the mass approaches 0? I just can't fathom how an object with no mass can even have an energy based on what I know (which is evidently not much).

 

[PeterDonis]

If the energy of a photon is

E = (hbar)(omega)

and the units of hbar are J*s (obviously), then how can a photon have no mass if a J is defined as 1 kg(m^2)/(s^2)?

The question of whether a photon has mass is independent of your choice of units. In fact, in quantum mechanics, we often use units in which $\hbar = 1$, so the units of energy are inverse length.

I just can't fathom how an object with no mass can even have an energy

Because it has momentum and therefore kinetic energy. The most general energy-momentum relation in relativity is $E^2 = p^2 c^2 + m^2 c^4$. The two terms can be thought of as kinetic energy (due to momentum) and rest energy. For a photon, $m = 0$, so the photon's energy is all kinetic energy, no rest energy.

 

[jaurandt]

The question of whether a photon has mass is independent of your choice of units. In fact, in quantum mechanics, we often use units in which $\hbar = 1$, so the units of energy are inverse length.
Because it has momentum and therefore kinetic energy. The most general energy-momentum relation in relativity is $E^2 = p^2 c^2 + m^2 c^4$. The two terms can be thought of as kinetic energy (due to momentum) and rest energy. For a photon, $m = 0$, so the photon's energy is all kinetic energy, no rest energy.

Thanks for your reply! Could you please describe to me how setting hbar to 1 gives it units of length^-1 ?

Also, this still confuses me even in this context because, as far as I know, momentum is mass*velocity. So I still can't get around the mass issue.

 

[PeterDonis]

Could you please describe to me how setting hbar to 1 gives it units of length^-1 ?

Actually, to get that you have to set $\hbar = 1$ and $c = 1$, i.e., both of those have no units. Then, since Planck's constant is energy times time, and it has no units, energy and time must have inverse units. And since $c = 1$, time and distance have the same units (because the ratio of distance to time must have no units), so energy must have units of inverse length.

as far as I know, momentum is mass*velocity

That's true for objects with nonzero rest mass, yes (and provided they are traveling at speeds small compared with the speed of light--otherwise you need to use the correct relativistic formula). But it's not the most general definition of momentum. The most general definition of momentum involves force applied over some period of time (another older word for this is "impulse", which you will still see in some contexts, such as rocketry). Photons can exert force (light pressure) on objects, and therefore they must have momentum.

 

[jaurandt]

I think I'm understanding. Basically if you take

E = (hbar)(k)(c)

And set **h** and c to 1, cancel the 2*pi's you're left with

E = 1/lamda

 

[PeterDonis]

Basically if you take

E = (hbar)(k)(c)

And set **h** and c to 1, cancel the 2*pi's you're left with

E = 1/lamda

Basically, yes.

 

[jaurandt]

Force applied over a period of time.

So that must mean

integral over t2 to t1 of force(t)dt
= momentum(t) |^t2_t1

How wrong is this?

 

[PeterDonis]

How wrong is this?

It's not wrong at all. The integral of force over time is the change in momentum over that time.